Problem: Let $f(x) = 2x^{2}-8x-4$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Answer: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $2x^{2}-8x-4 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 2, b = -8, c = -4$ $ x = \dfrac{+ 8 \pm \sqrt{(-8)^{2} - 4 \cdot 2 \cdot -4}}{2 \cdot 2}$ $ x = \dfrac{8 \pm \sqrt{96}}{4}$ $ x = \dfrac{8 \pm 4\sqrt{6}}{4}$ $x =2 \pm \sqrt{6}$